Question

A copper tube having 45mm internal diameter and 1.5mm wall thickness is closed at its ends by plugs which are at 450mm apart. The tube is subjected to internal pressure of 3 MPa and at the same time pulled in axial direction with a force of 3 kN. Compute: i) the change in length between the plugs ii) the change in internal diameter of the tube. Take Ecu = 100 GPa, and Mcu = 0.3.​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given A copper tube having 45 mm internal diameter and 1.5 mm wall thickness is closed at its ends by plugs which are at 450 mm apart. The tube is subjected to internal pressure of 3 MPa and at the same time pulled in axial direction with a force of 3 kN. Compute: i) the change in length between the plugs

• Now the copper tube is subjected to internal pressure of 3 MPa, Due to this
•                Longitudinal stress σ = p x d / 4 x t
•     p = internal pressure in the tube
• d = internal diameter of tube
• t = thickness of tube
• Length L = 450 mm
• So Longitudinal stress σ = 3 x 45 / 4 x 1.5
•                                        = 135 / 6
•                                        = 22.5 MPa
• So Longitudinal strain ε = p x d / 4 x t x (l - 2μ) / E
•                                       = 22.5 x (1 – 2 x 0.3) / 100 x 10^3
•                                       = 22.5 x 0.4 / 100 x 10^3
•                                      = 9/100 x 10^3
•                                         = 9 x 10^-5
• Now the change in length will be
•                                     ΔL = ε x L
•                                           = 9 x 10^-5 x 450
•                                           = 0.0405 mm
• So there is an increase.
• Now circumference strain ε = p x d / 4t (2 – μ) / E
•                                              = 22.5 x (2 – 0.3) / 100 x 10^3
•                                               = 38.25 / 100 x 10^3
•                                                = 0.3825 x 10^-3 m
• Now change in diameter will be  
•                                         Δd = ε x d
•                                               = 0.3825 x 10^-3 x 45
•                                               = 17.21 x 10^-3 m
•                                              = 0.01721 mm
• Now there is a force of 3KN in the axial dire x direction
• So Area of cross section will be A = π x d x t
•                                                        = 22/7 x 45 x 10^-3 x 1.5
•                                                       = 212.14 mm^2
• So for longitudinal strain ε = σ / E
•                                            = P/A x E
•                                                       = 3 x 10^3 / 212.14 x 100 x 10^3
•                                                        = 1.414 x 10^-4
• Now change in length
•                                     Δl = ε x L
•                                        = 0.06363 mm
• So there is an increase.
• So change in length due to combined effect will be
•                                       0.0405 + 0.06363
•                                          = 0.10413 mm

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